What is the integral of (1-e^{-x})/x ?

The integral of (1-e^{-x})/x is ln|x|-\Ei(-x)+C

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integral of (1-e^{-x})/x

integral

\frac{1 - e ^{- x}}{x}

ln ∣ x ∣ - Ei (- x) + C

Solution steps

\int \frac{1 - e ^{- x}}{x} d x

Expand

\frac{1 - e ^{- x}}{x} : \frac{1}{x} - \frac{e ^{- x}}{x}

= \int \frac{1}{x} - \frac{e ^{- x}}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x} d x - \int \frac{e ^{- x}}{x} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{e ^{- x}}{x} d x = Ei (- x)

= ln ∣ x ∣ - Ei (- x)

Add a constant to the solution

= ln ∣ x ∣ - Ei (- x) + C

\frac{x ^{3}}{( x + 1 ) ^{3}}

2^{x} \cdot e^{2} x

\frac{( x - 1 )}{( x ^{2} + 3 x - 4 )}

\frac{( x ^{3} - 1 )}{x ^{2}}

\frac{z}{( 1 - z ^{2} ) ^{\frac{1}{2}}}

What is the integral of (1-e^{-x})/x ?
The integral of (1-e^{-x})/x is ln|x|-\Ei(-x)+C